Question

A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is $5m$, the number of balls thrown per minute is $(takeg=10\frac{m}{s^2})$

• A. 30
• B. 60
• C. 90
• D. 45

Answer 1

Answer: (B)

60

Maximum height of ball = $5m$
So, velocity of projection $\Rightarrow u=\sqrt{2gh}=\sqrt{2\times 10\times 5}=10\frac{m}{s}$
time interval between two balls (time of ascent) $=\frac{u}{g}=1sec=\frac{1}{60}min.$
So no. of balls thrown per min = 60